poj1961

maksyuki 发表于 oj 分类，标签: 字符串-KMP

15 9月 2015

Period

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the
number zero on it.

Output

For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

aaa

aabaabaabaab

Sample Output

Test case #1

2 2

3 3

Test case #2

2 2

6 2

9 3

12 4

Source

Southeastern Europe 2004

题目类型：KMP中Next数组的应用

算法分析：直接求得输入字符串的Next数组，然后从小到大枚举字符串前缀子串的长度。对于每个前缀子串，判断i % (i – Next[i])是否为0且满足i/ (i – Next[i]) > 1，如果满足则输出前缀子串的长度和i / (i – Next[i])

#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <iomanip>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <map>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <list>
#include <ctime>
using namespace std;

const int INF = 0x7FFFFFFF;
const int maxn = 1000000 + 66;

int Next[maxn], len;

string s;

void Build ()
{
    Next[0] = -1, Next[1] = 0;
    int k = 0, i;

    for (i = 2; i <= len; i++)
    {
        while (k >= 0 && s[i-1] != s[k])
            k = Next[k];

        Next[i] = ++k;
    }
}

int main()
{
//    ifstream cin ("aaa.txt");

    int flag = 1;

    while (cin >> len)
    {
        if (len == 0)
            break;

        cin >> s;

        Build ();

        cout << "Test case #" << flag++ << endl;

        int i;
        for (i = 1; i <= len; i++)
        {
            if (i % (i - Next[i]) == 0 && i / (i - Next[i]) > 1)
                cout << i << " " << i / (i - Next[i]) << endl;
        }

        cout << endl;
    }

return 0;
}

#include <iostream>

#include <fstream>

#include <sstream>

#include <algorithm>

#include <iomanip>

#include <cstring>

#include <cstdio>

#include <cmath>

#include <map>

#include <string>

#include <vector>

#include <stack>

#include <queue>

#include <set>

#include <list>

#include <ctime>

using namespace std;

const int INF = 0x7FFFFFFF;

const int maxn = 1000000 + 66;

int Next[maxn], len;

string s;

void Build ()

{

Next[0] = -1, Next[1] = 0;

int k = 0, i;

for (i = 2; i <= len; i++)

{

while (k >= 0 && s[i-1] != s[k])

k = Next[k];

Next[i] = ++k;

}

int main()

{

// ifstream cin ("aaa.txt");

int flag = 1;

while (cin >> len)

{

if (len == 0)

break;

cin >> s;

Build ();

cout << "Test case #" << flag++ << endl;

int i;

for (i = 1; i <= len; i++)

{

if (i % (i - Next[i]) == 0 && i / (i - Next[i]) > 1)

cout << i << " " << i / (i - Next[i]) << endl;

}

cout << endl;

}

return 0;

}

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